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1000 dollars were converted into pounds and then the pounds

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1000 dollars were converted into pounds and then the pounds [#permalink] New post 27 Aug 2008, 02:07
1000 dollars were converted into pounds and then the pounds were converted back into dollars at the same exchange rate of x pounds per dollar. If a commission of y% is levied on any exchange operation, what dollar amount was left after the exchanges?

1. x = 0.6
2.y = 5


* Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
* Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
* BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
* EACH statement ALONE is sufficient
* Statements (1) and (2) TOGETHER are NOT sufficient
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Re: conversion [#permalink] New post 27 Aug 2008, 03:00
I pick B. If the exchange rate is the same and is applied two times on the opposite direction, it cancels itself - no need to know it.
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Re: conversion [#permalink] New post 27 Aug 2008, 05:41
Nerdboy wrote:
I pick B. If the exchange rate is the same and is applied two times on the opposite direction, it cancels itself - no need to know it.




I pick C.

Initial conversion
1000 $ = 1000*0.6 *1-0.05) = 600 =570 pounds
Reverse conversion
570 pounds= 570*1/0.6*0.95=902.5 $
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Last edited by x2suresh on 27 Aug 2008, 06:17, edited 1 time in total.
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Re: conversion [#permalink] New post 27 Aug 2008, 05:51
It should be "B". Same exchange rate will cancel out in both directions, so not to think of that, only the fee levied needs to be known.

To make it more clear;

$1000----> GBP results in 10x*(100-y)GBP

Again conversion of above amount back to Dollars results in $(100-y)^2/10

Only variable which needs to be known to know the amount of Dollar is y. Thus "B".
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Re: conversion [#permalink] New post 27 Aug 2008, 06:15
LM wrote:
It should be "B". Same exchange rate will cancel out in both directions, so not to think of that, only the fee levied needs to be known.

To make it more clear;

$1000----> GBP results in 10x*(100-y)GBP

Again conversion of above amount back to Dollars results in $(100-y)^2/10

Only variable which needs to be known to know the amount of Dollar is y. Thus "B".


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you are right..
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Re: conversion [#permalink] New post 27 Aug 2008, 07:45
Are you sure the exchange rate cancels itself out?

If I convert $100 into GBP, lets say I then have £200. If the % rate is 5%, then the first time, I get charged $5US, so I have $95 to convert and then I have £190. In the exchange back, the stem doesn't tell us the % is based upon one certain form of currency, so it seems that 5% would be levied upon the £190 as well, thus actually giving us £180.50 to exchange back into dollars. And if the exchange rate is 2:1 £:$, then £180.50 becomes $90.25. It doesn't cancel itself out. I'm not sure the wording of the question is exactly clear on how the commission % is calculated.

Either way you need to know the % commission otherwise in the exchange from £ back to $, you don't know how much is left after the first exchange, so you don't know how much is commission on the second exchange from £ to $.
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Re: conversion [#permalink] New post 27 Aug 2008, 08:00
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jallenmorris wrote:
Are you sure the exchange rate cancels itself out?

If I convert $100 into GBP, lets say I then have £200. If the % rate is 5%, then the first time, I get charged $5US, so I have $95 to convert and then I have £190. In the exchange back, the stem doesn't tell us the % is based upon one certain form of currency, so it seems that 5% would be levied upon the £190 as well, thus actually giving us £180.50 to exchange back into dollars. And if the exchange rate is 2:1 £:$, then £180.50 becomes $90.25. It doesn't cancel itself out. I'm not sure the wording of the question is exactly clear on how the commission % is calculated.

Either way you need to know the % commission otherwise in the exchange from £ back to $, you don't know how much is left after the first exchange, so you don't know how much is commission on the second exchange from £ to $.


We don't need exchange rate but we need commision %

1000 $
convert $ to £
= 1000 *x*
= first converted amount received + first commision paid
= 1000*x (1-y/100)+ 1000*x*y/1000
reconvert amount 1000*x (1-y/100) £ to $ ( commision levied is y)
= second converted amount received+ second commision paid
= [1000*x (1-y/100) ] *1/x(exchange rate) * (1-y/100) + [1000*x (1-y/100) ] *1/x * y/100

second converted amount received= 1000 * (1-y/100) ^2
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Re: conversion [#permalink] New post 27 Aug 2008, 17:09
can someone pls plug in and explain??
Re: conversion   [#permalink] 27 Aug 2008, 17:09
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